In the expansion of $$\left(9x - \frac{1}{3\sqrt{x}}\right)^{18}$$, $$x > 0$$, the term independent of $$x$$ is $${221} k$$. Then $$k$$ is equal to :

$$T_{r+1} = \binom{n}{r} a^{n-r} b^r$$

$$T_{r+1} = \binom{18}{r} (9x)^{18-r} \left(-\frac{1}{3}x^{-1/2}\right)^r$$

$$T_{r+1} = \binom{18}{r} \cdot 9^{18-r} \cdot (-1)^r \cdot \left(\frac{1}{3}\right)^r \cdot x^{18-r} \cdot x^{-r/2}$$

$$T_{r+1} = \binom{18}{r} \cdot (3^2)^{18-r} \cdot (-1)^r \cdot (3^{-1})^r \cdot x^{18 - r - \frac{r}{2}}$$

$$T_{r+1} = \binom{18}{r} \cdot 3^{36 - 3r} \cdot (-1)^r \cdot x^{18 - \frac{3r}{2}}$$

$$18 - \frac{3r}{2} = 0 \implies 18 = \frac{3r}{2} \implies r = \frac{18 \times 2}{3} = 12$$ (term independent of $$x$$)

$$T_{13} = \binom{18}{12} \cdot 3^{36 - 3(12)} \cdot (-1)^{12}$$

$$T_{13} = \binom{18}{6} \cdot 3^0 \cdot 1 = \binom{18}{6}$$

$$221k = 221 \times 84 \implies k = 84$$